3A — f x »* * + 3 A + 2B—2 x n + A + B + C — | 
= 0; whence (by the lemma) sA — I = 0, 3A +- 
2B — 2 — 0, and A + B f C — 1—0; therefore 
l 
(f ; 
n 2 v z 
*nd consequently \ + 4 + 0 + id .»* = — + -r 
3 ^ 
Case3°, To determine the sum o f the progression i 3 4 tv 
H- 3 J + 4 3 n 3 , or 1 f 8 + 27 + 64 n 3 . 
By putting An* + B>* 3 4- C/? 1 f D?t zr l + 8 -+ 27 
+ 64 M 3 , and proceeding as above, we shall have 
4An 3 4- 6A« 1 f 4An + A + 3B?i* + 3B/i + B + 2Cn 
+ C + D ( — n -f f| 3 ) zr n 3 + 3n z + 3n f l ; and 
therefore ~4A — i x n 3 f 6A + 3B — 3 x n z + 
3 X n + A + B . 4- C f D — l —0; 
3B, 
4 A + 3B + 2C 
hence A= - 1 -, B 
4 v 3 
6A _ ] r ,,_3—4 A ■ 
~ "2 ’ ^ 2 
71* — 
+ 
+ 
or 
4 ' 2 ' 4 
In the very same manner it will 
found that 
l 4 + 2 + f 3* 
I s 4- 2 5 + 3 s 
l 6 + 2 6 + 3" 
, IV ir , n i n 
. = — 4 + . 
5 2 3 30 
. , Jl $ , 5 H* 71* 
* ~ 6 + 2 + 12 12* 
« 7l 7 . 7I 6 . n 5 « 3 , 77 
,71 2= ' ~ + -g- 4 g- + —» 
&’C. 
In order to exemplify what l^as been thus for deliver 
ed, let it, in the lirst place, be required to find the sum 
of the series of squares 1 + 4 + 0 f 16, &c. continued 
to 10 terms; then by substituting 10 for n, in thege- 
&c.